Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
51.
The HCF (GCD) of a, b is 12, a, b are positive integers and a > b > 12. The smallest values of (a, b) are respectively ?
(A) 12, 34
(B) 24, 12
(C) 24, 36
(D) 36, 24
Solution:
HCF (GCD) of a, b number is 12 and a > b > 12 (given) ∴ Smallest value of a & b are (36, 24)
52.
What is the greatest number of 3 digits which when divided by 6, 9 and 12 leaves a remainder of 3 in each case ?
(A) 903
(B) 939
(C) 975
(D) 996
Solution:
Greatest number of three digits is 999. LCM of 6, 9 and 12 is 36 On dividing 999 by 36, the remainder obtained is 27. So, required number = (999 - 27) + 3 = 975
53.
Find the greatest number of five digits which when divided by 3, 5, 8, 12 leaves 2 as remainder ?
(A) 99999
(B) 99948
(C) 99962
(D) 99722
Solution:
LCM of (3, 5, 8, 12) ⇒ 3 × 5 × 4 × 2 = 120 ⇒ Now greatest five digits number is 99999 On dividing 99999 by = 120 (LCM) We get remainder = $$\frac{{99999}}{{120}},$$ remainder = 39 ⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12) ∴ 99999 - 39 = 99960 ⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case. ⇒ add 2 in the 99960 ⇒ 99960 + 2 ⇒ 99962
54.
252 can be expressed as a product of primes as:
(A) 2 × 2 × 3 × 3 × 7
(B) 2 × 2 × 2 × 3 × 7
(C) 3 × 3 × 3 × 3 × 7
(D) 2 × 3 × 3 × 3 × 7
Solution:
Clearly, 252 = 2 × 2 × 3 × 3 × 7
55.
The ratio of two numbers is 3 : 4 and their LCM is 120. The sum of numbers is = ?
(A) 70
(B) 140
(C) 35
(D) 105
Solution:
Let the numbers are 3x and 4x So HCF = x ∴ HCF × LCM = Product of numbers ⇒ x × 120 = 3x × 4x ⇒ x × 120 = 12x2 ⇒ 120 = 12x x = 10 ∴ Numbers are 30 and 40 ∴ Sum of two numbers = 30 + 40 = 70
56.
The least number which when divided by 6, 9, 12, 15, 18 leaves the same remainder 2 in each case is = ?
(A) 180
(B) 176
(C) 182
(D) 178
Solution:
LCM of 6, 9, 12, 15, 18 is = 180 If 180 is divided by these given number remainder will be 0 ⇒ To leave the same remainder 2 ⇒ The number will be = 180 + 2 = 182
57.
L.C.M. of two numbers is 2079 and their H.C.F. is 27. If one of the number is 189, the other number is = ?
(A) 297
(B) 584
(C) 189
(D) 216
Solution:
Given, H.C.F. = 27 L.C.M. = 2079 one number = 189 Let another number be y ⇒ Product of numbers = L.C.M. × H.C.F. ∴ 189 x y = 27 × 2079 y = 297
58.
The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is = ?
(A) 4203
(B) 4023
(C) 4032
(D) 4302
Solution:
LCM of 12, 18, 21, 32 = 2016 = 2016 × K = 2016 × 2 = 4032 (K = 2) "4032" is the number which is completely divided by 12, 18, 21, 32
59.
The LCM of two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one number is 93, find the other = ?
The number of pairs of positive integers whose sum is 99 and HCF is 9 is = ?
(A) 5
(B) 2
(C) 3
(D) 4
Solution:
According to question, HCF = 9 ⇒ Then the two number will be 9a, 9b ⇒ 9a + 9b = 99 ⇒ a + b = 11 ⇒ Pairs of positive integer (1, 10) (2, 9) (3, 8) (4, 7) (5, 6) = 5